1. Formal methods in software development
a.y.2017/2018
Prof. Anna Labella
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2. see dens.pdf
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10.  fi()  g Proof  f()  f2()  f3()…..........  fi()
f ( fi() =  fi+1()
Given another fixpoint g, we have the constant chain
 g  g  g…  g = g
greater than the first one step by step, hence
 fi()  g
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11. We could start from the maximum and compute the
maximal fixpoint using the duality of order
We can use minimum/ maximum fixed point to solve
recursive equations like
Ex: x=ax+b
If we do not have inverse operations, we have to
use approximation as we do in the case of formal
languages Ø
{b}
{ab, b}
{a2b, ab, b}.....
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12. Fixed point theorem: an example
A fixed point x for a function f:(S)(S) is an element of (S) such that f(x) = x
We will give an interpretation of CTL operators using fixed points
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13. Fixed point theorem: an example
Let S be a set (of states) of a transition system K and
f:(S)(S) a monotonic function w.r.t.  , then f has a minimal and a maximal fixpoint
nfn(Ø) and nfn(S), respectively
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14. Fixed point theorem (Tarski)
Let S be a set (of states) and
f:(S)(S) a monotonic function w.r.t.  ,
Starting from S the maximal fixpoint of f is: S
f(S)
f2(S)
nfn(S)
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15. Looking for a semantic for CTL
We have to find the set of states satisfying a formula φ,
symbolically, SK(φ) or |[φ]|.
Let us start with a simple example:
we are given with a transition system K
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20. Recursively defined operators
AG φ = φ∧AXAG φ
EG φ = φ∧EXEG φ
AF φ = φ∨AXAF φ
EF φ = φ∨EXEF φ
A[φ U ψ] = ψ ∨( φ∧AXA[φ U ψ])
E[φ U ψ] = ψ ∨( φ∧EXE[φ U ψ])
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21. The example again
We want to define the set of states where EGx is the greatest solution of the recursive equation x = x∧ EX (x)
i.e. as the maximal fixpoint of the operator p = ( )∧ EX ( ) : (S)(S)
We start from the greatest subset S and define pre(S’) as the subset of states such that there is for them a successor state in S’
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26. An example again
We want to define the set of states where E(xUy) is the smallest solution of the recursive equation <x,y> = y ∨(x∧ EX (x,y))
i.e. as the minimal fixpoint of the operator
x = ( )∧ EX( ) : (S)(S)
We start from the smallest subset of S, namely Ø
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